Set Difference Then Union Equals Union Then Set Difference
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Theorem
Let $S, A, B$ be sets.
Let $A \cap B = \O$.
Then:
- $\paren {S \setminus A} \cup B = \paren {S \cup B} \setminus A$
Corollary
Let $S, T$ be sets.
Let $A \subseteq S \setminus T$.
Let $B \subseteq T \setminus S$.
Then:
- $\paren{S \setminus A} \cup B = \paren{S \cup B} \setminus A$
Proof
We have:
\(\ds \paren {S \cup B} \setminus A\) | \(=\) | \(\ds \paren {S \setminus A} \cup \paren {B \setminus A}\) | Set Difference is Right Distributive over Union | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {S \setminus A} \cup B\) | Set Difference with Disjoint Set |
$\blacksquare$